AFRICAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Imhotep Mathematical Proceedings
Volume 4, Numéro 1, (2017),
pp. 34 - 43.
On counting subgroups for a class of finite nonabelian p-groups
and related problems
Olapade Oluwafunmilayo O.
Department of Mathematics,
University of Ibadan, Ibadan, Oyo
State, Nigeria. Abstract
ola1182006@gmail.com The main goal of this article is to review the work of Marius Tărnăuceanu,
where an explicit formula for the number of subgroups of finite nonabelian
M. Enioluwafe p-groups having a cyclic maximal subgroups was given. Using examples to
clarify our work and in addition we give an explicit formula to some related
Department of Mathematics, problems.
University of Ibadan, Ibadan, Oyo
State, Nigeria.
michael.enioluwafe@gmail.com
Proceedings of the 6th annual workshop on
CRyptography, Algebra and Geometry http://imhotep-journal.org/index.php/imhotep/
(CRAG-6), 15 - 17 June 2016, University of Imhotep Mathematical Proceedings
Bamenda, Bamenda, Cameroon.
IBADAN UNIVERSITY LIBRARY
IMHOTEP - Math. Proc. 4 (2017), 34–43 IMHOTEP Mathematical
1608-9324/010034–33, DOI 13.1007/s00009-003-0000
©c 2017 Imhotep-Journal.org/Cameroon Proceedings
On counting subgroups for a class of finite nonabelian
p-groups and related problems.
Olapade Oluwafunmilayo O. and EniOluwafe M.
Abstract. The main goal of this article is to review the work of Marius Tărnăuceanu, where
an explicit formula for the number of subgroups of finite nonabelian p-groups having a cyclic
maximal subgroups was given. Using examples to clarify our work and in addition we give
an explicit formula to some related problems.
Mathematics Subject Classification (2010). 20D60 (primary), 20D15, 20D30, 20D40. (sec-
ondary).
Keywords. Finite nonabelian p-groups, Cyclic subgroups, Number of subgroups, Recurrence
relation, Cartesian products.
I. Preliminaries
Counting subgroup of finite groups is one of the most important problems of combinatorial
finite group theory. Starting with the last century, this topic has enjoyed a steady and gradual
process of development. The problem was completely solved in the abelian case, by establishing
an explicit expression of the number of subgroups of a finite abelian group (see[2]). Several
authors have worked on this area using different methods: Gautomi Bhowmik [2] used Gauss-
ian polynomial to evaluate divisor function of matrices, Călugăceanu G [3] and J Petrillo [6]
used Goursat’s lemma for groups to derive explicit formulae, Marius Tărnăuceanu [10] and
EniOluwafe M. [4] used the concept of fundamental group lattice to count some types of sub-
groups of a finite nonabelian group; Tărnăuceanu in [11] used method based on certain attached
matrix, Lászlo T óth [7] and Amit Sehgal [1] use simple group-theoretic and number theoretic
formulae. Unfortunately, in the nonabelian case such expression can be given only for few classes
of finite groups.
In the following let p be a prime, n ≥ 3 be an integer and consider the class G of all finite
nonabelian p-group of order p possessing a maximal subgroup which is cyclic. A detailed de-
scription of G is given by Theorem 4.1, chapter 4, [8]: a group is contained in the class G if and
only if it is isomorphic to
n−1 n−2
M(pn) =< x, y | xp = yp = 1, y−1xy = xp +1 > when p is odd, or to one of the next groups
- M(2n) (n ≥ 4),
- the dihedral group
D =< x, y | x2n−1 n−12n = y2 = 1, yxy−1 = x2 −1 >
Communication presentée au 6ème atelier annuel sur la CRyptographie, Algèbre et Géométrie (CRAG-6), 15 - 17 Juin
2016, Université de Bamenda, Bamenda, Cameroun / Paper presented at the 6th annual workshop on CRyptography,
Algebra and Geometry (CRAG-6), 15 - 17 June 2016, University of Bamenda, Bamenda, Cameroon.
IBADAN UNIVERSITY LIBRARY
Vol. 4 (2017) Counting subgroups for a class of finite nonabelian p-groups 35
- the generalized quaternion group
n−1 n−1
Q2n =< x, y | x2 = y4 = 1, yxy−1 = x2 −1 >
- the quasidihedral group
n−1 n−2
S2n =< x, y | x2 = y2 = 1, y−1xy = x2 −1 >
(n ≥ 4)
when p = 2. If G is a group, then the set L(G) consisting all subgroups of G forms
a complete lattice with respect to set inclusion, called the subgroup lattice of G. Most of our
notation is standard and will usually not be repeated here. For basic definition and results on
groups we refer the reader to [9] and [8]. In this paper we use examples to make the work of
Marius more explicit. In his work he determine the cardinality of L(G) for the groups G in G,
by using the above presentation and their main properties (collected in (4.2), chapter 4, [8]).
II. Main results
II.1. Modular groups
First of all, T ănăuceanu [10] find the number of subgroups of Modular group M(pn). And state
some of the property of the Modular groups:
• The commutator subgroup D(M(pn)) has order p and is generated by xq, where q = pn−2.
• Ω (M(pn1 )) =< xq, y >∼= Zp × Zp.
• M(pn) contains p+ 1 minimal subgroups.
• The join of any two distinct minimal subgroups includes D(M(16)).
n | 2n−1 n−2Let p = 2, then M(2 ) =< x, y x = y2 = 1, y−1xy = x2 +1 >,
We give the following examples to make the above properties more explicit:
II.1.1. Example. when n = 4,
M(16) =< x, y | x8 = y2 = 1, y−1xy = x3 >
= {1, x, x2, x3, x4, x5, x6, x7, y, xy, x2y, x3y, x4y, x5y, x6y, x7y}
q 22D(M(16)) =< x >=< x >=< x4 >= {1, x4}
Clearly, D(M(16)) has order 2.
II.1.2. Example. Let Ω1(M(16)) =< x
4, y >= {1, x4, y, x4y}, so Ω1(M(16)) has order 4
Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)}
Z2 × Z2 has order 4.
We compute this table to see clearly the structure of Ω1(M(16)) and Z2 × Z2.
Clearly from the table, Ω (M(16)) ∼1 = Z2 × Z2.
II.1.3. Example. Minimal subgroups of M(16) are:
{1, x4}, {1, y}, {1, x4y}
Clearly, M(16) contains 3 minimal subgroups which is p+ 1.
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36 Olapade Oluwafunmilayo O. and EniOluwafe M.
Table 1. Analysis of order of elements of Ω1(M(16)) and Z2 × Z2
Order of elements 1 2 4
Ω1(M(16)) 1 x
4, y x4y
Z2 × Z2 (0, 0) (0, 1), (1, 0) (1, 1)
Total number 1 2 1
II.1.4. Example. The join of any two distinct minimal subgroups include D(M(pn)): Join
{1, x4} and {1, y} gives {1, x4, y, x4y}
1, x4 ∈ {1, x4, y, x4y}
Clearly it includes D(M(16)).
From the above, the following results were obtained:
| M(p
n)
L(M(pn)) |=| L( ) | +p+ 1. (1)
D(M(pn))
recall that the commutator subgroup, D(M(pn)) is a minimal subgroup and that’s the reason
for adding p+ 1 in equation (1) above.
II.1.5. Example. recall: GH = {gH | g ∈ G},
M(16)
=< x, y >
D(M(16))
M(16)
= {gD(M(16)) | g ∈M(16)}
D(M(16))
= {D(M(16)), xD(M(16)), x2D(M(16)), x3D(M(16)), yD(M(16)),
xyD(M(16)), x2yD(M(16)), x3yD(M(16))}
M(pn)
D(M(pn)) is an abelian group.
II.1.6. Example. M(16)D(M(16)) is abelian if xyD(M(16)) = yxD(M(16)).
yxD(M(16)) = x5yD(M(16)) (yx = x5y)
= x · x4yD(M(16)) (x · x4y = x5y)
= xyx4D(M(16)) (x4y = yx4)
= xyD(M(16)) (x4 ∈ D(M(16)))
∴ yxD(M(16)) = xyD(M(16)) (that is x commutes with y)
Hence M(16)D(M(16)) is abelian.
M(pn)
D(M(pn)) is of order p
n−1,
that is:
M(16)
D(M(16)) is of order 8
|M(pn)|
|D(M(pn))| = |
M(16) 16
D(M(16)) | = 2 = 8.
This is confirmed by the number of elements in M(16)D(M(16)) (example 2.1.5)
n
Next, we show that M(p ) =∼n Zp × Zpn−2D(M(p )) have isomorphic lattices of subgroups. Thus, we
need to determine the number of subgroups of certain order for Zp×Z M(16)pn−2 and that of D(M(16)) .
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Vol. 4 (2017) Counting subgroups for a class of finite nonabelian p-groups 37
II.1.7. Example. Let p = 2, n = 4, we have:
Z ∼2 × Z4 = Z8
Z2 = {1, a}
Z4 = {1, y, y2, y3}
Z2 × Z4 = {(1, 1), (1, y), (1, y2), (1, y3), (x, 1), (x, y), (x, y2), (x, y3)}
Z2 × Z4 is of order 8.
Likewise,
M(16)
= {D(M(16)), xD(M(16)), x2D(M(16)), x3D(M(16)), yD(M(16)),
D(M(16))
xyD(M(16)), x2yD(M(16)), x3yD(M(16))}
M(16)
D(M(16)) Is also of order 8.
We compute these tables to see clearly the structure of M(16)D(M(16)) and Z2 × Z4.
Table 2. Analysis of order of elements of Z2 × Z4
Order of elements 1 2 4
Z2 × Z4 (1, 1) (1, y), (1, y2), (x, 1) (1, y3), (x, y), (x, y2), (x, y3)
Total number 1 3 4
Table 3. Analysis of order of elements of M(16)D(M(16))
Order of elements 1 2 4
M(16)
D(M(16)) (1, 1) x
2D(M(16)), yD(M(16)), x2yD(M(16)) xD(M(16)), x3D(M(16)), xyD(M(16)), x3yD(M(16))
Total number 1 3 4
Comparing the order of M(16)D(M(16)) and Z2×Z4 and the order of their elements (as shown
on the tables 2 and 3 above), we conclude that they are isomorphic. Therefore,
M(pn)
=∼ Zp × Z n−2 (2)
D(M(pn)) p
n
Being isomorphic, the groups M(p )D(M(pn)) and Zp × Zpn−2 have isomorphic lattices of subgroups.
Thus, their is a need to determine the number of subgroups of Zp × Zpn−2 . In order to do this
he recall the following auxiliary result, established in [11, Theorem 3.3, pp.378].
Lemma 1. For every 0 ≤ α ≤ α + α , the number of all subgroups of order pα1+α2−α1 2 in the
finite abelian p− group Zpα1 × Zpα2 (α 1
≤ α2) is:
 α+1
p − 1
, if 0 ≤ α ≤ α
− 1p 1
α
p 1
+1 − 1
 , if α1 ≤ α ≤ α− 2 p 1pα1+α2−α+1 − 1 , if α2 ≤ α ≤ α− 1 + α2.p 1
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38 Olapade Oluwafunmilayo O. and EniOluwafe M.
In particular, the total number of subgroups of Zpα1 × Zpα2 is:
1
[(α − α + 1)pα1+2 − (α − α − 1)pα1+12 1 2 1 − (α1 + α2 + 3)p+ (α1 + α2 + 1)]
(p− 2)2
For α1 = 1 and α2 = n− 2, it results:
| 1L(Zp × Zpn−2)| = [(n− 2)p3 − (n− 4)p2 − (n+ 2)p+ n)] = (n− 2)p+ n. (3)(p− 2)2
Now, the relation (1), (2) and (3) show that the next theorem holds.
Theorem 2. The number of subgroups of the group M(pn) is given by the following equality:
|L(M(pn))| = (n− 1)p+ n+ 1.
Proof. Recall from [1] that
n
| L(M(pn | | M(p ))) = L( ) | +p+ 1
D(M(pn))
and from [2] that
M(pn) ∼= Zp × Zpn−2D(M(pn))
and from [3]
| M(p
n)
L(M(pn)) | =| L( ) | +p+ 1
D(M(pn))
= Zp × Zpn−2 + p+ 1
= (n− 2)p+ n+ p+ 1
= (n− 2 + 1)p+ n+ 1
= (n− 1)p+ n+ 1
Hence, | L(M(pn)) |= (n− 1)p+ n+ 1

Next, we focus on the groups D2n , Q2n and SD2n . An important property of these groups is
that their centres are of order 2 (they are generated by xq, where q = 2n−2) Marius [10] gave
the properties and we cite examples for clarity. That is, Z (D2n),Z (Q2n) and Z (SD2n) are of
order 2 and are generated by < xq >
Example. when n = 4, p = 2
Z (D 42n) = Z (D16) = {1, x }
Z (Q2n) = Z (Q16) = {1, x4}
Z (D2n) = Z (SD16) = {1, x4}
when n = 5, p = 2
Z (D2n) = Z (D32) = {1, x8}
Z (Q 82n) = Z (Q32) = {1, x }
Z (D2n) = Z (SD32) = {1, x8}
For any G ∈ {D2n , Q2n , SD2n} we have:
G ∼= D n−1 (4)
Z (G) 2
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II.2. Dihedral groups
Let n = 4, G = D16, Z(D16) = {1, x4}
G
= {gZ(G)|g ∈ G}
Z(G)
D16
= {gZ(D16)|g ∈ D16}
Z(D16)
D16 = {1, x, x2, x3, x4, x5, x6, x7, y, xy, x2y, x3y, x4y, x5y, x6y, x7y}
D16 is of order 16
D16
= {Z(D ), xZ(D ), x216 16 Z(D16), x3Z(D ), yZ(D ), xyZ(D ), x2yZ(D ), x316 16 16 16 yZ(D16)}
Z(D16)
D16
Z(D is of order 816)
D2n−1 = D8 = {1, x, x2, x3, y, xy, x2y, x3y} which is of order 8.
For D2n this isomorphism will lead us to a recurrence relation verified by | L(D2n) |, but first
we need to compute the number of subgroups in D2n which does not contain Z (D2n)(that is the
number of subgroups of D2n nZ (D ) ). Clearly, the trivial subgroup of D2 as well as all its minimal2n
subgroup excepting Z (D2n) (that are of the form < x
iy >, i = 0, 2n−1 − 1) satisfy this property.
Since for every i 6= j = 0, 2n−1 − 1 we have xiyxjy = xi−j .
II.2.1. Example. xiyxjy = xi−j .
x2yx3y = x2x5yy = x7 = x−1 (yx3 = x5y;x−1 = x7)
x4yx2y = x4x6yy = x10 = x2 (x8 = 1)
x5yx2y = x5x6yy = x3 (yx2 = x6y)
Table 4. Analysis of order of elements of D2n−1
Order of elements 1 2 4
D 22n−1 1 x , y, xy, x
2y, x3y x, x3
Total number 1 5 2
Table 5. Analysis of order of elements of D16Z(D16)
Order of elements 1 2 4
D16 2
Z(D 6) Z(D16) x Z(D16), yZ(D16), xyZ(D16), x
2yZ(D16), x
3yZ(D16) xZ(D
3
16), x Z(D16)
1
Total number 1 5 2
Considering the equality of the order of elements and the order of the groups above
(as we can see in table 3 and 4), we can conclude that they have the same structure and are
isomorphic.
It follows again that the join of any two distinct minimal subgroups in D2n includes
Z (D2n).
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Table 6. Analysis of the number of subgroups in D2n
D2n Order1 Order2 Order4 Order8 Order16 Order32 Order64 | L(D2n−1) | Formula
D8 1 5 3 1 – – 10 – 2
3 + 2
D16 1 9 5 3 1 – 19 – 2
4 + 3
D32 1 17 9 5 3 1 – – 2
5 + 4
.. .. .
. .
. .. . .
. .. .. .. . . . ..
D 1 2(n−1)−1 + 1 2(n−1)−2 + 1 2(n−1)−3 + 1 2(n−1)−42n−1 + 1 – – · · · 2n−1 + (n− 2)
II.2.2. Example. Joining {1, y} and {1, x2y} gives {1, x2, x4, x6, y, x2y, x4y, x6y} and {1, x4} ∈
{1, x2, x4, x6, y, x2y, x4y, x6y}
So, by a similar reasoning as for M(pn), we obtain that the number of subgroups of D2n verifies
the recurrence relation
n
|L(D2n)| |
D2
= L( )|+ 2n−1 + 1
Z(D2n)
|L(D2n)| = |L(D n−12n−1)|+ 2 + 1. (5)
for all n ≥ 3. Writing (5) for n = 3, 4, ... and |L(D2n−1)| is 2n−1n− 2 (from table [3]. Summing
up these equalities, we find an explicit expression of |L(D2n)|.
Theorem 3. The number of subgroups of the group D2n is given by the following equality:
|L(D2n)| = 2n + n− 1.
Proof. From (5) |L(D )| = |L(D )|+2n−1+1. From table (6) |L(D )| is 2n−12n 2n−1 2n−1 n−2
then,
|L(D2n)| = 2n−1 + (n− 2) + 2n−1 + 1.
= 2 · 2n−1 + (n− 2) + 1.
= 2 · 2n−1 + n− 1.
= 2n + n− 1.

II.3. Quaternion groups
Because Q2n verifies also the relation (4) and Z(Q2n) is the unique minimal subgroup of Q2n ,
we can easily infer from Theorem 3.
Theorem 4. The number of subgroups of the group Q2n is given by the following equality:
| L(Q2n) | =| L(D2n−1) | +1
= 2n−1 + (n− 1)− 1 + 1
= 2n−1 + n− 1
II.4. Quasi-dihedral groups(SD2n)
The method developed above can also be used to count the subgroups of the quasi-dihedral
group (SD2n)n ≥ 4. For each i ∈ 0, 1, . . . , 2n−1 − 1, we have (xiy)2 = xiq. Hence ord(xiy) = 2
when i is even, while ord(xiy) = 4 when i = odd. This shows that the minimal subgroups of
S2n are of the form < x
q > and < x2jy >, j = 0, 2n−2 − 1.
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II.4.1. Examples. For each i ∈ 0, 1, . . . , 2n−1 − 1
• (xiy)2 = xiq For n = 4, i = 3, q = 2n−2
(xiy)2 = (x3y)2
= x3yx3y
= x3xyy(xy = yx3)
= x4
xiq = x3·4
= x12
= x8 · x4
= x4
Clearly, (xiy)2 = xiq
• ord(xiy) = 2 when i is even Let i = 2
(x2y)2 = x2y · x2y
= x2x6yy(x6y = yx2)
= x8y2
= 1
Clearly when i is even xiy is of order two.
• ord(xiy) = 4 when i is odd Let i = 3
(x3y)4 = (x3y)2 · (x3y)2 = x4 · x4
= x8
= 1
Clearly when i is odd xiy is of order four.
• Minimal subgroups are of the form < xq > and < x2jy >,
n = 4, q = 2n−2, j = {0, 1, . . . , 2n−2 − 1}
For SD16 we have:
{1, x4} of the form < xq >.
and
{1, y}, {1, x2y}{1, x4y}, {1, x6y} of the form < x2jy > which is 4 in number.
Clearly for SD16 we have 5 minimal subgroup.
Let n = 5,
For SD32 we have:
{1, x8} of the form < xq >,
and
{1, y}, {1, x2y}, {1, x4y}, {1, x6y}, {1, x8y}, {1, x10y}, {1, x12y}, {1, x14y} of the form< x2jy >
which is 8 in number, that is, 23
Clearly for SD32 we have 9 minimal subgroup.
It is clear that the minimal subgroup without the centre can be written as a power of
prime, and of this form: 2n−2.
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42 Olapade Oluwafunmilayo O. and EniOluwafe M.
The join of any two distinct minimal subgroups different from < xq > contains a nonzero power
of x and therefore it includes < xq >.
II.4.2. Example. Combining 1, y and 1, x2y we have {1, x2, x4, x6, y, x2y, x4y, x6y} and {1, x4} ∈
{1, x2, x4, x6, y, x2y, x4y, x6y}. Thus we conclude that the subgroups of SD2n which does not
contain Z(S2n) are:
n−1
< 1 >,< y >,< x2y >, . . . , < x2 −2 > .
In view of the group isomorphism SD2n ∼ n−1Z(SD ) = D2 , which gives2n
|L(SD n−22n)| = |L(D2n−1)|+ 2 + 1, (6)
for all n ≥ 4. From (6) and theorem 3 we get immediately the next result.
Theorem 5. | L(SD2n) |= 3 · 2n−2 + n− 1,
Proof. Recall from table 7 that
|L(D2n−1)| = 2n−1n− 2
| SD2
n
L(SD2n) | =| L | +2n−2 + 1
Z(SD2n)
=| L(D ) | +2n−22n−1 + 1
= 2n−1 + n− 2 + 2n−2 + 1
= 2n−1 + 2n−2 + n− 1
= 2 · 2n−2 + 2n−2 + n− 1
= 3 · 2n−2 + n− 1

Finally, for an arbitrary finite group it is not an easy task comparing the number of its subgroups
and the number of its elements. But can be easily made for the 2-groups in our class G, by using
Theorems 3, 4, and 5. Obviously, it obtains:
| L(M(2n)) |≤|M(2n) |, for all n ≥ 3
| L(D2n) |>| D2n |, for all n ≥ 3
| L(Q2n) |<| Q2n |, for all n ≥ 3
| L(SD2n) |<| SD2n |, for all n ≥ 4
Moreover, the following limits were calculated:
|L(D2n)|
lim = 1
n→∞ |D2n |
|L(Q2n)| 1
lim =
n→∞ |Q2n | 2
|L(SD2n)| 3
lim = .
n→∞ |SD2n | 4
For any fixed prime p, we also have:
|L(Mpn)|
lim = 0
n→∞ |Mpn |
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Vol. 4 (2017) Counting subgroups for a class of finite nonabelian p-groups 43
III. Related Problems
Arising from this work are other related problems which we are working on. One of the problem
is given below:
III.1. Counting Subgroups of the groups of type: D2n × C2
D2n is a dihedral group of order 2
n, n ≥ 3, and C2 is a cyclic group of order 2.
Table 7. Analysis of the number of subgroups in D2n × C2
D2n Order1 Order2 Order4 Order8 Order16 Order32 Order64 | L(D2n × C2) | Formula
D8 × C2 1 11 15 7 1 – −− 35 25 + 3(1)
D16 × C2 1 19 27 15 7 1 – 70 26 + 3(2)
D32 × C2 1 35 51 27 15 7 1 137 27 + 3(3)
.. .. .. .. .. .. .. .D2n × C2 . . . . . . . .. 2n+2 + 3(n− 2)
Theorem 6. For n ≥ 3, the number of subgroups of the group D2n ×C2 is given by the following
equality:
| L(D2n × C2) |= 2n+2 + 3(n− 2)
Where | L(D2n × C2) | is the subgroup lattice of D2n × C2.
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Olapade Oluwafunmilayo O.
e-mail: ola1182006@gmail.com
EniOluwafe M.
e-mail: michael.enioluwafe@gmail.com
Department of Mathematics, Faculty of Science, University of Ibadan, Ibadan, Oyo State, Nigeria.
Imhotep Proc.
IBADAN UNIVERSITY LIBRARY